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Difference between revisions of "2013 AMC 10B Problems/Problem 24"

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Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.
 
Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.
  
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Factoring 2016, we get <math>2^5*3^2*7</math>. <math>wlog</math>, we have <math>a<b</math>.
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Factoring 2016, we get <math>2^5*3^2*7</math>. WLOG, we have <math>a<b</math>.
  
 
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.
 
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.

Revision as of 18:26, 30 December 2017

Problem

A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$) such that the sum of the four divisors is equal to $n$. How many numbers in the set $\{ 2010,2011,2012,\dotsc,2019 \}$ are nice?


$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$

Solution 1

A positive integer with only four positive divisors has its prime factorization in the form of $a*b$, where $a$ and $b$ are both prime positive integers or $c^3$ where $c$ is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of $a*b$. The four factors of this number would be $1$, $a$, $b$, and $ab$. The sum of these would be $ab+a+b+1$, which can be factored into the form $(a+1)(b+1)$. Easily we can see that now we can take cases again.

Case 1: Either $a$ or $b$ is 2.

If this is true then we have to have that one of $(a+1)$ or $(b+1)$ is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either $\frac{2016}{3} - 1$ or $\frac{2010}{3} - 1$ is a prime. We see that in this case none of them work.

Case 2: Both $a$ and $b$ are odd primes.

This implies that both $(a+1)$ and $(b+1)$ are even which implies that in this case the number must be divisible by $4$. This leaves only $2012$ and $2016$. $2012=4*503$ so either $(a+1)$ or $(b+1)$ both has a factor of $2$ or one has a factor of $4$. If it was the first case, then $a$ or $b$ will equal $1$. That means that either $(a+1)$ or $(b+1)$ has a factor of $4$. That means that $a$ or $b$ is $502$ which isn't a prime, so 2012 does not work. $2016 = 4 * 504$ so we have $(503 + 1)(3 + 1)$. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is $\boxed{\textbf{(A)}\ 1}$.

Solution 2

If $m$ has four divisors, then its divisors would be 1, $a$, $b$ and $ab$, where $a$ and $b$ are prime. Therefore, the sum of the divisors of $m$ is $1+a+b+ab=(a+1)(b+1)$.

If either $a+1$ or $b+1$ are odd, then $a$ or $b$ are even. Therefore, $a+1$ and $b+1$ are even, so $m$ is a multiple of 4. The only two numbers from the range $2010-2019$ that are multiples of 4 are 2012 and 2016.

Factoring 2012, we get $2^2*503$. To make $a+1$ and $b+1$ even, $wlog$, we have $a+1=2$ and $b+1=1006$. However, if $a$ was 1, then $a$ is not prime, so 2012 is not nice.

Factoring 2016, we get $2^5*3^2*7$. WLOG, we have $a<b$.

Testing for the lowest $a$, we get $a+1=4$ and $b+1=504$. Therefore, $a=3$, and $b=503$, so $n=2016$ is nice, with $m=1509$. Therefore, the answer is $\boxed{\textbf{(A)}\ 1}$.

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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